Find all values of x so that the triangle with vertices A=(−4, −3, x), B=(−7, 0, 5), and C=(−1, −6, 8) has a right angle at A.
as with the other question just like this one, you want AB•AC = 0
So,
A=(−4, −3, x), B=(−7, 0, 5), and C=(−1, −6, 8)
(-7+4)(-1+4) + (0+3)(-6+3) + (5-x)(8-x) = 0
x = 2 or 11
Hello, can you help me with this question?
Thanks oobleck, I just figured it out lol... thanks for confirming I got it right
can someone pls help me
To determine whether a triangle has a right angle at a specific vertex, we can check if the dot product of the vectors formed by that vertex with the other two vertices is zero. In this case, we need to find all values of x such that the dot product of vectors AB and AC is zero.
Let's start by calculating the vectors AB and AC:
Vector AB = B - A
= (-7, 0, 5) - (-4, -3, x)
= (-7, 0, 5) + (4, 3, x)
= (-7 + 4, 0 + 3, 5 + x)
= (-3, 3, 5 + x)
Vector AC = C - A
= (-1, -6, 8) - (-4, -3, x)
= (-1, -6, 8) + (4, 3, x)
= (-1 + 4, -6 + 3, 8 + x)
= (3, -3, 8 + x)
Now, let's calculate the dot product of AB and AC:
AB ⋅ AC = (-3)(3) + (3)(-3) + (5 + x)(8 + x)
= -9 - 9 + (40 + 8x + 5x + x^2)
= -18 + 40 + 8x + 5x + x^2
= x^2 + 13x + 22
To find the values of x for which the dot product is zero, we need to solve the equation:
x^2 + 13x + 22 = 0
We can factor this quadratic equation or use the quadratic formula to find the solutions. Factoring might not be straightforward in this case, so let's use the quadratic formula:
x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1, b = 13, and c = 22. Plugging these values into the formula, we get:
x = (-13 ± √(13^2 - 4(1)(22))) / (2(1))
x = (-13 ± √(169 - 88)) / 2
x = (-13 ± √81) / 2
x = (-13 ± 9) / 2
This gives us two possible values of x:
1. x = (-13 + 9) / 2 = -2
2. x = (-13 - 9) / 2 = -11/2 = -5.5
Therefore, the triangle with vertices A=(-4, -3, x), B=(-7, 0, 5), and C=(-1, -6, 8) has a right angle at A when x equals -2 or -5.5.